YES

We show the termination of the TRS R:

  .(.(x,y),z) -> .(x,.(y,z))
  a(f(x)) -> f(a(x))
  a(.(x,y)) -> .(a(x),y)
  a(b1(x)) -> b1(a(x))
  f(b(x)) -> b(f(x))
  .(b(x),y) -> b(.(x,y))
  b1(b(x)) -> b(b(x))
  a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
  a(f(|0|())) -> b1(.(f(|0|()),|0|()))
  f(.(|0|(),x)) -> b(.(|0|(),f(x)))
  f(|0|()) -> b(|0|())
  c(b(x)) -> c(a(x))
  a(b(x)) -> b(a(x))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: .#(.(x,y),z) -> .#(x,.(y,z))
p2: .#(.(x,y),z) -> .#(y,z)
p3: a#(f(x)) -> f#(a(x))
p4: a#(f(x)) -> a#(x)
p5: a#(.(x,y)) -> .#(a(x),y)
p6: a#(.(x,y)) -> a#(x)
p7: a#(b1(x)) -> b1#(a(x))
p8: a#(b1(x)) -> a#(x)
p9: f#(b(x)) -> f#(x)
p10: .#(b(x),y) -> .#(x,y)
p11: a#(f(.(|0|(),x))) -> b1#(.(f(.(|0|(),x)),.(|0|(),f(x))))
p12: a#(f(.(|0|(),x))) -> .#(f(.(|0|(),x)),.(|0|(),f(x)))
p13: a#(f(.(|0|(),x))) -> .#(|0|(),f(x))
p14: a#(f(.(|0|(),x))) -> f#(x)
p15: a#(f(|0|())) -> b1#(.(f(|0|()),|0|()))
p16: a#(f(|0|())) -> .#(f(|0|()),|0|())
p17: f#(.(|0|(),x)) -> .#(|0|(),f(x))
p18: f#(.(|0|(),x)) -> f#(x)
p19: c#(b(x)) -> c#(a(x))
p20: c#(b(x)) -> a#(x)
p21: a#(b(x)) -> a#(x)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The estimated dependency graph contains the following SCCs:

  {p19}
  {p4, p6, p8, p21}
  {p1, p2, p10}
  {p9, p18}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: c#(b(x)) -> c#(a(x))

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r13

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = ((1,1),(1,0)) x1
        b_A(x1) = ((1,1),(0,0)) x1 + (1,1)
        a_A(x1) = ((1,1),(0,0)) x1 + (1,1)
        ._A(x1,x2) = x1
        f_A(x1) = ((1,1),(0,0)) x1 + (2,1)
        b1_A(x1) = ((1,1),(0,0)) x1 + (1,1)
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = ((1,1),(0,0)) x1
        b_A(x1) = x1 + (1,1)
        a_A(x1) = ((0,1),(1,0)) x1
        ._A(x1,x2) = x1
        f_A(x1) = x1
        b1_A(x1) = ((1,1),(1,1)) x1
        |0|_A() = (0,0)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a#(b(x)) -> a#(x)
p2: a#(b1(x)) -> a#(x)
p3: a#(.(x,y)) -> a#(x)
p4: a#(f(x)) -> a#(x)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a#_A(x1) = ((1,1),(0,1)) x1
        b_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        b1_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a#_A(x1) = ((0,1),(1,1)) x1
        b_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        b1_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        ._A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: .#(.(x,y),z) -> .#(x,.(y,z))
p2: .#(b(x),y) -> .#(x,y)
p3: .#(.(x,y),z) -> .#(y,z)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The set of usable rules consists of

  r1, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        .#_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2
        ._A(x1,x2) = x1 + x2 + (1,0)
        b_A(x1) = ((1,1),(0,0)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        .#_A(x1,x2) = ((0,1),(0,1)) x1
        ._A(x1,x2) = ((0,1),(0,1)) x1 + x2 + (1,1)
        b_A(x1) = x1 + (1,2)
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(b(x)) -> f#(x)
p2: f#(.(|0|(),x)) -> f#(x)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = ((0,1),(1,1)) x1
        b_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        ._A(x1,x2) = x1 + ((0,1),(1,1)) x2 + (0,1)
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = (0,0)
        b_A(x1) = (1,1)
        ._A(x1,x2) = x1 + (0,1)
        |0|_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.